Electric field outside a long wire carrying charge q is proportional to :
$\frac{1}{r^{2}}$
$\frac{1}{r}$
$\frac{1}{r^{\frac{1}{3 / 2}}}$
$\frac{1}{r^{\frac{1}{3 / 5}}}$

Question

Electric field outside a long wire carrying charge q is proportional to -dfrac - 1 - - r - 2 - - dfrac - 1 - r - dfrac - 1 - - r - frac - 1 - 3-2 - - - dfrac - 1 - - r - frac - 1 - 3-5 - - -

Toppr Admin · Accepted Answer

Using Gauss Law across the cylindrical Gaussian surface,
$\int E . d S = \frac{q_{e n c}}{ϵ_{0}}$
$∴ E \times 2 π r \times l = \frac{λ \times l}{ϵ_{0}}$
$∴ E = \frac{λ}{2 π r ϵ_{0}}$
Here $λ$ is the linear charge density on the wire.

Electric field outside a long wire carrying charge q is proportional to :1r21r1r13/21r13/5

Using Gauss Law across the cylindrical Gaussian surface,∫E.dS=qencϵ0∴E×2πr×l=λ×lϵ0∴E=λ2πrϵ0Here λ is the linear charge density on the wire.

Electric field outside a long wire carrying charge q is proportional to :
$\frac{1}{r^{2}}$
$\frac{1}{r}$
$\frac{1}{r^{\frac{1}{3 / 2}}}$
$\frac{1}{r^{\frac{1}{3 / 5}}}$

Using Gauss Law across the cylindrical Gaussian surface,
$\int E . d S = \frac{q_{e n c}}{ϵ_{0}}$
$∴ E \times 2 π r \times l = \frac{λ \times l}{ϵ_{0}}$
$∴ E = \frac{λ}{2 π r ϵ_{0}}$
Here $λ$ is the linear charge density on the wire.